What volume of water results from complete combustion of a $40 \cdot L$ $40*L$ volume of methane under standard conditions?

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What volume of water results from complete combustion of a $40 \cdot L$ $40*L$ volume of methane under standard conditions?

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Answer 1 · 2017-05-05T17:54:59

The problem with this question is that we must ASSUME standard conditions........

$C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)$ $CH_4(g) + 2O_2(g) rarr CO_2(g) +2H_2O(l)$

Explanation:

Methane is a room temperature gas...., $boiling point =$ $"boiling point "=$ $- 164$ $-164$ $^{\circ} C$ $""^@C$ . And of course water is a room temperature liquid.....

We assume that the methane gas is under standard conditions, i.e. $298 \cdot K$ $298*K$ , and $1 \cdot a t m$ $1*atm$ .

And this represents $\frac{40 \cdot L}{24.5 \cdot L \cdot m o l^{- 1}} = 1.63 \cdot m o l$ $(40*L)/(24.5*L*mol^-1)=1.63*mol$ of the hydrocarbon.

Given the stoichiometry, we get................

$\frac{2 \times 1.63 \cdot m o l \times 18.01 \cdot g \cdot m o l^{- 1}}{1.0 \cdot g \cdot m L^{- 1}}$ $(2xx1.63*molxx18.01*g*mol^-1)/(1.0*g*mL^-1)$

$≅ 60 \cdot m L$ $~=60*mL$

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What volume of water results from complete combustion of a $40 \cdot L$ $40*L$ volume of methane under standard conditions?

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What volume of water results from complete combustion of a 40⋅L40*L volume of methane under standard conditions?

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Explanation:

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What volume of water results from complete combustion of a $40 \cdot L$ $40*L$ volume of methane under standard conditions?